Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
 
class Main {
    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int N = Integer.parseInt(br.readLine());
        
        String []s = br.readLine().split(" ");
 
 
        double []pi = new double[N];

        for(int i=0; i < N; ++i)
            pi[i] = Double.parseDouble(s[i]);


        double [][]dp = new double[N+1][N+1];

        for(int i=0; i <= N; ++i)
            for(int j=0; j < i+1; ++j) {
                if (i == 0)
                    dp[i][j] = 1.0;
                else if (j == 0)
                    dp[i][j] = dp[i-1][j] * (1.0 - pi[i-1]);
                else
                    dp[i][j] = dp[i-1][j] * (1.0 - pi[i-1]) + dp[i-1][j-1] * pi[i-1];
            }

        double ans = 0;

        for(int i = (N+1)/2; i <= N; ++i)
            ans += dp[N][i];
        System.out.println(ans);
    }
}